## Intermediate Algebra for College Students (7th Edition)

The solution set is {$\frac{3-\sqrt {5}}{4}, \frac{3+\sqrt {5}}{4}$}.
$x^{-2} - 6x^{-1} = -4$ Let $u=x^{−1}$ Rewrite the given equation as a quadratic equation in $u$ and solve for $u$. $x^{-2} - 6x^{-1} = -4$: $u^2 - 6u = -4$ Solve for $u$ by completing the square. Add $(\frac{-6}{2})^2 = 9$ to both sides. $u^2 - 6u +9= -4+9$ $u^2 - 6u +9= 5$ $(u -3)^2= 5$ Apply the square root property. $u -3= ±\sqrt {5}$ $u = 3±\sqrt {5}$ Solve for $u$. $u=x^{−1}$ $3±\sqrt {5}=x^{−1}$ $3±\sqrt {5}=\frac{1}{x}$ $x= \frac{1}{3±\sqrt {5}}$ $x= \frac{1}{3+\sqrt {5}}$ or $x= \frac{1}{3-\sqrt {5}}$ $x= \frac{3-\sqrt {5}}{4}$ or $x= \frac{3+\sqrt {5}}{4}$ The solution set is {$\frac{3-\sqrt {5}}{4}, \frac{3+\sqrt {5}}{4}$}.